Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $t \neq 0$. $a = \dfrac{4t + 16}{-3t + 6} \times \dfrac{-4t^2 + 68t - 280}{t^2 - 3t - 28} $
First factor out any common factors. $a = \dfrac{4(t + 4)}{-3(t - 2)} \times \dfrac{-4(t^2 - 17t + 70)}{t^2 - 3t - 28} $ Then factor the quadratic expressions. $a = \dfrac {4(t + 4)} {-3(t - 2)} \times \dfrac {-4(t - 7)(t - 10)} {(t - 7)(t + 4)} $ Then multiply the two numerators and multiply the two denominators. $a = \dfrac {4(t + 4) \times -4(t - 7)(t - 10) } {-3(t - 2) \times (t - 7)(t + 4) } $ $a = \dfrac {-16(t - 7)(t - 10)(t + 4)} {-3(t - 7)(t + 4)(t - 2)} $ Notice that $(t - 7)$ and $(t + 4)$ appear in both the numerator and denominator so we can cancel them. $a = \dfrac {-16\cancel{(t - 7)}(t - 10)(t + 4)} {-3\cancel{(t - 7)}(t + 4)(t - 2)} $ We are dividing by $t - 7$ , so $t - 7 \neq 0$ Therefore, $t \neq 7$ $a = \dfrac {-16\cancel{(t - 7)}(t - 10)\cancel{(t + 4)}} {-3\cancel{(t - 7)}\cancel{(t + 4)}(t - 2)} $ We are dividing by $t + 4$ , so $t + 4 \neq 0$ Therefore, $t \neq -4$ $a = \dfrac {-16(t - 10)} {-3(t - 2)} $ $ a = \dfrac{16(t - 10)}{3(t - 2)}; t \neq 7; t \neq -4 $